[Doi 10.1017_CBO9781316536483.004] Apte, Shaila Dinkar -- Signals and Systems (Principles and Applications) __ CT and DT Systems

3 CT and DT Systems Learning Objectives

ƒ Properties of systems

ƒ Invertibility

ƒ Linearity ƒ Time invariance

ƒ Memory ƒ LTI systems

ƒ Stability

ƒ Systems as interconnected operators

ƒ Causality

ƒ Series and parallel interconnection

This chapter concentrates on system definition and properties. The properties of systems such as linearity, time invariance, causality, invertibiity, memory and stability are discussed in detail. We have discussed sampling in chapter 1. The reader is already familiar with CT and DT signals. We will define properties of CT and DT systems simultaneously. Finally, we will see how to represent a system as an interconnection of operators and series/parallel; connection of systems.

3.1

Properties of CT and DT Systems – Linearity and Shift Invariance We will discuss and explain properties of CT and DT systems in the following sections. We will study the property of linearity and Shift/time invariance in this section.

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3.1.1

Linearity property Any CT or DT system is said to be linear if it obeys two important properties. The first is homogeneity and the second is additivity. Homogeneity property

The system is said to obey the property of homogeneity if the following condition holds:

• For CT and DT input signal of x(t) or x[n], respectively, if the output is given by y(t) or y[n], then if the input signal is scaled by a factor of k to get input signal of kx(t) or kx[n], then the output is also scaled by the same factor k i.e., output is ky(t) or ky[n], where k is any scaling factor. Let H represent the system operator, then

• if y(t ) = H[x(t )] then ky(t ) = H[kx(t )] and

(3.1)

• if y[n] = H[x[n]] then ky[n] = H[kx[n]]

(3.2)

Signals and Systems

Additivity Property (Superposition Property)

182

The system is said to obey additivity property if the following condition holds good.

• Let y1(t) and y1[n] be the output for the CT and DT input signal x1(t) and

x1[n], respectively, and y2(t) and y2[n] be the output for input signal x2(t) and x2[n], respectively, for a CT and DT system. The property of additivity states the following: If the input to a system is the addition of two signals, then the output of the system is the addition of the respective outputs. Let H represent the system operator, then

if y1 (t ) = H[x1 (t )]and y2 (t ) = H[x2 (t )],then • If then y1 (t ) + y2 (t )= H[x (t ) + x2 (t )]

(3.3)

if y1[n] = H[x1[n]]and y2 [n] = H[x2 [n]],then then If y1[n] + y2 [n=] H[x [n] + x2 [n]]

(3.4)

This indicates that the signals added at the input do not interfere with each other when they pass via a system. Considering homogeneity and additivity holding good, we can write the following statement for CT and DT systems. This is a statement of superposition property. All the linear systems obey the superposition property.

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if y1 (t ) = H[x1 (t )]and y2 (t ) = H[x2 (t )],then then • If ay1 (t ) + by2 (t= ) H[ax (t ) + bx2 (t )]

(3.5)

if y1[n] = H[x1[nt ]]and y2 [n] = H[x2 [n]],then then If ay1[n] + by2 [n=] H[ax [n] + bx2 [n]]

(3.6)

Physical significance of linearity

Check if the homogeneity property holds good for a system given by y(t) = 5? Find if the system is linear.

CT and DT Systems

If the system is linear, the transfer graph of output Vs input is a straight line graph passing through the origin. Linear systems have a meaning more than this! If the system is linear, the input signal can be suitably decomposed into component signals and the corresponding outputs for the component signals one at a time can be calculated by assuming all other inputs equal to zero. The component outputs can be scaled and added to generate the output of the system for the input signal. This procedure simplifies the computation of the system output for different input signals. This property is called as the superposition property. We will consider some examples of CT and DT systems to illustrate the concept of linearity. We will first consider CT systems

Solution

183

Example 3.1

Note that for any input, the output is constant. We will have to check if the homogeneity and additivity property hold good. We will first check homogeneity property. If the input is doubled (scaled by a factor of 2), the output is not scaled by the same factor. It remains constant equal to 5. According to the property of homogeneity, if y(t ) = H[x(t )] then 2 y= (t ) H[2 × x(t )]

(3.7)

Here, the output remains constant even if the input is doubled. Hence, the system is not homogeneous. We will now check for additivity. If the input to the system is x1(t) + x2(t), then the output must be y1 (t ) + y2 (t ) = 5 + 5 = 10

(3.8)

if the property of additivity holds good. The actual output is just equal to 5 whatever the applied input is. The additivity also fails. The system is not homogeneous and not additive and hence, is not linear. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Example 3.2

Check if the system given by y(t) = x(t) + 2 is linear. Solution

In the given system, for an input x(t), the output is given by y(t) = x(t) + 2

(3.9)

Therefore for the input kx(t), where k is any scaling factor, according to homogeneity property, the output would be y(t) = k[x(t) + 2] ≠ k[x(t)] + 2

(3.10)

Hence, the system is not homogeneous. We will now check for additivity. If the input to the system is x1(t) + x2(t), then output must be

Signals and Systems

y1 (t ) + y2 (t ) = x1 (t ) + x2 (t ) + c .

184

(3.11)

The actual output is equal to y1 (t ) + y2 (= t ) x1 (t ) + c + x2 (t ) + c

(3.12)

The additivity property also fails. Teaser The graph for both above systems is a straight line, yet they are nonlinear systems! The above systems are said to have a bias. The systems with bias are incrementally linear. The incrementally linear system will have a graph of Δy Vs Δx as linear and it will pass through the origin. Example 3.3

Check if the system given by y(t) = x(t) + x(t – 1) is linear. Solution

Let the system be y(t) = x(t) + x(t – 1). Suppose the input is kx(t), where k is any scaling factor. Then the output is y(t) = kx(t) + kx(t – 1)

(3.13)

To satisfy the homogeneity property the output should be y(t) = k[x(t) + x(t – 1)], which is the actual output. Hence, the system is homogeneous. To satisfy the property of additivity, if the input to the system is x1(t) + x2(t) then output must be y1 (t ) + y2 (t= ) x1 (t ) + x1 (t − 1) + x2 (t ) + x2 (t − 1).

(3.14)

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The actual output is equal to y1 (t ) + y2 (t= ) x1 (t ) + x1 (t − 1) + x2 (t ) + x2 (t − 1)

(3.15)

The additivity property holds good. Hence, the system is linear. Note

The system obeys superposition property and hence, is linear!

Example 3.4

Find if the system given by y(t) = 0.1 × x(t) is linear? Solution

Let the system be y(t) = 0.1 × x(t). Suppose the input is kx(t), where k is any scaling factor. Then the output is y(n) = 0.1k × x(t)

(3.16)

y(t ) =k × 0.1 × x(t ) =0.1kx(t )

(3.17)

Hence, the system is homogeneous. To satisfy the property of additivity, if the input to the system is x1(t) + x2(t), then the output must be y1 (t ) + y2 (t ) = 0.1 ×[x1 (t )] + 0.1 ×[x2 (t )].

(3.18)

CT and DT Systems

To satisfy the homogeneity property the output should be

185

The actual output is equal to y1 (t ) + y2 (t ) = 0.1 ×[x1 (t ) + x2 (t )]

(3.19)

The additivity property also holds good. Hence, the system is linear. Note The graph of the system is linear and it passes through the origin! Let us solve some examples for DT systems. Example 3.5

Does the homogeneity property hold for y[n] = 2? Is the system linear? Solution

Note that for any input, the output is constant. We will have to check if the homogeneity and additivity property hold good. We will first check the homogeneity property. If the input is doubled (scaled by a factor of 2), the output is not scaled by the same factor. It remains constant equal to 2. Hence, the system is not homogeneous. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

We will now check for additivity. If the input to the system is x1[n] + x2[n], then the output must be y1[n] + y2 [n] = 2 + 2 = 4

(3.20)

if the property of additivity holds good. The actual output is just equal to 2 whatever the applied input is. The additivity also fails. The system is not homogeneous and not additive and hence, is not linear. Example 3.6

Consider the following system y[n] = x[n] + c. Is the system homogeneous? Is it linear? Solution

In the given system, for an input x(n) the output is given by

Signals and Systems

y[n] = x[n] + c

186

(3.21)

Therefore for the input kx[n], where k is any scaling factor, the output would be y[n] = k[x[n]] + c

(3.22)

According to the condition of homogeneity property the output in this case should be k times the output when the input was x[n], that is, the output should be k[x[n] + c]

(3.23)

which is not the case and hence, the system is not homogeneous. We will now check for additivity. If the input to the system is x1[n] + x2[n], then the output must be y1[n] + y2 [n] = x1[n] + x2 [n] + c.

(3.24)

The actual output is just equal to y1[n] + y2 [n=] x1[n] + c + x2 [n] + c

(3.25)

The additivity property also fails. ✓ Teaser The graph for both the systems is a straight line, yet they are nonlinear systems! The above systems are said to have a bias. The systems with bias are incrementally linear. The incrementally linear system will have a graph of Δy Vs Δx as linear and it will pass through the origin. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

✓ Things to remember The system is a linear system if the output Vs input graph for a system is linear and it passes the through origin. If the graph is linear and it does not pass through the origin, it is an incrementally linear system, but not a linear system. A system with a non-linear transfer graph is linear if it obeys the superposition principle. Example 3.7

Consider the following system y[n] = |x[n]|. Is the system linear? Solution

Suppose the input is kx[n], where k is any scaling factor, then the output is y[n] = |kx[n]|

(3.26)

To satisfy the homogeneity property the output should be (3.27)

Hence, the system is not homogeneous. The system is not homogeneous, therefore it is non-linear. To satisfy the property of additivity, if the input to the system is x1[n] + x2[n], then the output must be y1[n] + y2 [n] = | x1[n]| + | x2 [n]|.

(3.28)

187

The actual output is just equal to y1[n] + y2 [n] =| x1[n] + x2 [n]| .

CT and DT Systems

y[n] = k|x[n]|

(3.29)

The additivity property also fails. Hence, the system is non-linear. The transfer graph of the system is shown in Fig. 3.1.

Fig. 3.1 The transfer graph for the DT system shown as a continuous graph Let us now consider some examples of linear systems. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Example 3.8

Consider the following systems and verify the homogeneity and additivity property. 1. y[n] = 0 2. y[n] = x[0] 3. y[n] = x[n – 1] 4. y[n] = 2x[n] + 3x[n – 1]

Signals and Systems

5. y[n] = 0.2x[n] – 0.1y[n – 1]

188

Solution 1. Let the system be y[n] = 0. Suppose the input is kx[n], where k is any scaling factor, then the output is zero = ky[n]. To satisfy the homogeneity property the output should be y[n]= k × 0 = 0. Hence, the system is homogeneous. To satisfy the property of additivity, if the input to the system is x1[n] + x2[n], then the output must be

y1[n] + y2 [n] = 0 + 0 = 0.

(3.30)

The actual output is equal to y1[n] + y2 [n] = 0.

(3.31)

The additivity property also holds good. Hence, the system is linear. Note

The graph of the system is linear and it passes through the origin!

2. Let the system be y[n] = x[0]. Suppose the input is kx[n], where k is any scaling factor, then the output is just x[0]. To satisfy the homogeneity property the output should be y[n] = k × x[0]. Hence, the system is not homogeneous. To satisfy the property of additivity, if the input to the system is x1[n] + x2[n], then the output must be y1[n] + y2 [n] = x1[0] + x2 [0].

(3.32)

The actual output is equal to y1[n] + y2 [n] = x1[0] + x2 [0]

(3.33)

The additivity property holds good. Hence, the system is not linear. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Note The graph of the system is linear; however, it does not pass through the origin! 3. Let the system be y[n] = x[n – 1]. Suppose the input is kx(n), where k is any scaling factor. Then the output is y[n] = kx[n – 1]

(3.34)

To satisfy the homogeneity property the output should be y[n] = k × x[n – 1]. Hence, the system is homogeneous. To satisfy the property of additivity, if the input to the system is x1[n] + x2[n], then the output must be y1[n] + y2 [n=] x1[n − 1] + x2 [n − 1].

(3.35)

The actual output is equal to (3.36)

The additivity property holds good. Hence, the system is linear. Note The graph of the system is linear and it does pass through the origin! 4. Let the system be y([n] = 2x[n] + 3x[n – 1]. Suppose the input is kx[n], where k is any scaling factor, then the output is y([n] = 2kx[n] + 3kx[n − 1].

CT and DT Systems

y1[n] + y2 [n=] [x1[n − 1] + x2 [n − 1]]

189

(3.37)

To satisfy the homogeneity property the output should be y([n] = 2kx[n] + 3kx[n − 1]

(3.38)

Hence, the system is homogeneous. To satisfy the property of additivity, if the input to the system is x1[n] + x2[n], then the output must be y1[n] + y2 [= n] [2 x1[n] + 3x1[n − 1]] + [2 x2 [n] + 3x2 [n − 1]].

(3.39)

The actual output is the same. The additivity property also holds good. Hence, the system is linear. Note The system obeys superposition property and hence, is linear. 5. Let the system be y[n] = 0.2x[n] – 0.1y[n – 1]. Suppose the input is kx[n], where k is any scaling factor, then the output is Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

y[n] = k × (0.2 x[n] − 0.1 y[n − 1])

(3.40)

To satisfy the homogeneity property the output should be y[n] = k × 0.2 x[n] − k × 0.1 y[n − 1]

(3.41)

Hence, the system is homogeneous. To satisfy the property of additivity, if the input to the system is x1[n] + x2[n], then the output must be y1[n] + y= 2 [n] [0.2 x1[n] − 0.1 y1[n − 1)] + [0.2 x 2 [n] − 0.1 y 2 [n − 1)] .

(3.42)

The actual output is the same. The additivity property also holds good. Hence, the system is linear. Note The system obeys superposition property and hence, is linear.

Signals and Systems

Example 3.9

Consider the CT systems given by 1. y(t) = cos(x(t)) 2.

y(t ) = ∫

3.

y(t ) =

t /4

−∞

190

x(τ )dτ

d x(t ) dt

4. y(t) = x(3 – t) 5. y(t) = x(t/5) Are the systems linear? If not, explain why. Solution

1. Consider a system given by y(t) = cos(x(t)) Let the output for the inputs x1(t) and x2(t) be y1(t) and y2(t) given by y1(t) = cos(x1(t)) and y2(t) = cos(x2(t)), respectively. We will verify if the superposition property holds good. Let the input to the system be ax1(t) + bx2(t), then the output will be cos(ax1 (t )) + cos(bx2 (t )) ≠ a cos(x1 (t )) + b cos(x2 (t ))

(3.43)

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Hence, the system is not linear. We know that the cosine curve is non-linear! The system using any trigonometric function is non-linear!! 2.

y(t ) = ∫

t /4

−∞

x(τ )dτ

Let output for the inputs x1(t) and x2(t) be y1(t) and y2(t) given by t /4 t /4 y1 (t ) = ∫ x1 (τ )dτ , y2 (t ) = ∫ x 2 (τ )dτ , respectively. −∞

−∞

We will verify if the superposition property holds good. Let the input be ax1(t) + bx2(t), the output will be ay1 (t ) + by2 = (t ) a ∫

t /4

−∞

3.

y(t ) =

t /4

−∞

x2 (τ )= dτ

∫

t /4

−∞

[ax1 (τ ) + bx2 (τ )]dτ (3.44)

The system obeys superposition property and hence, is linear. d x(t ). Let us verify the superposition property. dt

Let output for the inputs x1(t) and x2(t) be y1(t) and y2(t) given by d d y1 (t ) = x1 (t ) , y2 (t ) = x2 (t ) , respectively. dt dt We will verify if the superposition property holds good. Let the input be ax1(t) + bx2(t), the output will be ay1 (t ) + by2 (t= ) Note

d d d (ax1 (t ) + bx2 (t )) = a x1 (t ) + b x2 (t ) dt dt dt

CT and DT Systems

Note

x1 (τ )dτ + b ∫

191

(3.45)

The system obeys superposition property and hence, is linear.

4. y(t) = x(3 – t) Let output for the inputs x1(t) and x2(t) be y1(t) and y2(t) given by y1(t) = x1(3 – t), y2(t) = x2(3 – t), respectively. We will verify if the superposition property holds good. Let the input be ax1(t) + bx2(t), the output will be ay1 (t ) + by2 (t= ) ax1 (3 − t ) + bx2 (3 − t ) Note

(3.46)

The system obeys superposition property and hence, is linear.

5. y(t) = x(t/5) Let output for the inputs x1(t) and x2(t) be y1(t) and y2(t) given by y1(t) = x1(t/5), y2(t) = x2(t/5), respectively. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

We will verify if the superposition property holds good. Let the input be ax1(t) + bx2(t), the output will be ay1 (t ) + by2 (t )= ax1 (t / 5) + bx2 (t / 5) Note

(3.47)

The system obeys superposition property and hence, is linear.

Example 3.10

Consider the DT systems given by 1. y[n] = sin(x[n]) 2. y[n] = 3x[n]u[n] 3. y[n] = log10(|x[n]|)

Signals and Systems

y[n] 4. =

192

n

∑ x[k + 1]

k = −∞

∞

5. y[n] x[n] ∑ δ [n − 3k] = k = −∞

6. y[n] = x[n3] Are the systems linear? If not, explain why. Solution

1. y[n] = sin(x[n]) Let output for the inputs x1[n] and x2[n]be y1[n] and y2[n] given by y1[n] = sin(x1[t]) and y2[n] = sin(x2[n]), respectively. We will verify if the superposition property holds good. Let the input be ax1[n] + bx2[n], the output will be sin(ax1[n]) + sin(bx2 [n]) ≠ a sin(x1[n]) + b sin(x2 [n])

(3.48)

Hence, the system is not linear. We know that the cosine curve is non-linear! The system using any trigonometric function is non-linear!! 2. y[n] = 3x[n]u[n] Let output for the inputs x1[n] and x2[n] be y1[n] and y2[n] given by y1[n] =3x1[n]u[n] and y2[n] =3x2[n]u[n], respectively. We will verify if the superposition property holds good. Let the input be ax1[n] + bx2[n], the output will be 3ax1[n]u[n] + 3bx2 [n]u[n]= [(a × 3(x1[n] + b × 3(x2 [n]))u[n]

(3.49)

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Hence, the system is linear. Note The system obeys superposition property and hence is linear. 3. y[n] = log10(|x[n]|) Let output for the inputs x1[n] and x2[n] be y1[n] and y2[n] given by y1[n] = log10(|x1[n]|) and y2[n] = log10(|x2[n]|), respectively. We will verify if the superposition property holds good. Let the input be ax1[n] + bx2[n], the output will be a log10 x1[n] + b log10 x2 [n] ≠ log10 [(a × (x1[n] + b × (x2 [n]))

(3.50)

Hence, the system is non-linear. Note The system does not obey superposition property and the transfer curve of log function is not linear; hence, the system is non-linear. n

∑ x[k + 1].

k = −∞

Let output for the inputs x1[n] and x2[n] be y1[n] and y2[n] given by = y1[n]

n

∑

k = −∞

y2 [n] x1[k + 1] and=

n

∑ x [k + 1], respectively.

k = −∞

2

We will verify if the superposition property holds good. Let the input be ax1[n]+bx2[n], the output will be n

n

a ∑ x1[k + 1] + b ∑ x2 [= k + 1] k = −∞

k = −∞

n

∑ ax [k + 1] + bx [k + 1]

k = −∞

1

2

(3.51)

CT and DT Systems

y[n] 4. =

193

Hence, the system is linear. Note The system obeys superposition property and hence, is linear. ∞

= 5. y[n] x[n] ∑ δ [n − 3k] k = −∞

Let output for the inputs x1[n] and x2[n] be y1[n] and y2[n] given by ∞

∞

k = −∞

k = −∞

y1[n] = x1[n] ∑ δ [n − 3k] and y2 [n] = x2 [n] ∑ δ [n − 3k] , respectively. We will verify if the superposition property holds good. Let the input be ax1[n] + bx2[n], the output will be ∞

∞

∞

k = −∞

k = −∞

k = −∞

ax1[n] ∑ δ [n − 3k] + bx2 [n] ∑ δ [n − 3k=] [ax1[n] + bx2 [n]] ∑ δ [n − 3k]

(3.52)

Hence, the system is linear. Note The system obeys superposition property and hence, is linear. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

6. y[n] = x[n2] Let output for the inputs x1[n] and x2[n] be y1[n] and y2[n] given by y1[n] = x1[n2] and y2[n] = x2[n2], respectively. We will verify if the superposition property holds good. Let the input be ax1[n] + bx2[n], the output will be ax1[n2 ] + bx2 [n2 ] =ay1[n] + by2 [n]

(3.53)

Hence, the system is linear. Note The system obeys superposition property and hence, is linear.

Signals and Systems

3.1.2

194

Time invariance / shift invariance property Most signal-processing (DSP) techniques also require the system to be shift invariant even though it is not a requirement for linearity. The system is said to be time invariant or shift invariant if its input–output characteristics do not change with time. If y(t) or y[n] is the output for some input x(t) or x[n], respectively, for analog and digital systems, then for a shift invariant system, if the input is shifted by k time units / k samples to the right, that is, the input is x(t – k)/x[n – k] then the output is y(t – k)/y[n – k], that is, the output is also shifted by k time units/k samples to the right. The system that is linear and obeys the time invariance property is called a linear time invariant (LTI) system. We will refer to the LTI systems only unless stated otherwise. Physical significance of time/shift invariance If the system is time/shift

invariant, then the output of the system for any shifted input can be easily calculated by introducing same amount of shift in the output.

If the system is LTI, one can characterize the system in terms of its impulse response. One may decompose any input signal into scaled and shifted delta functions and the output of the system can be obtained using the principle of superposition. The output of the system to a shifted delta function is just the shifted impulse response. The computation of the output to any input signal is greatly simplified. We will illustrate the property of shift invariance with the help of CT and DT systems.

Physical significance of linear time/shift invariance

Example 3.11

(Non-linear and non shift invariant system) Is the following system linear and shift/time invariant? y(t) = x(1). Solution

We will verify the shift invariance property. Let the input be shifted by k time units, therefore the new input is x(t – k). The output is still x(1) and is not Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

shifted by k time units. So, the system is not a shift variant system. We will now check for linearity. If the input is kx(t), the output must be kx(1). However, the actual output is just x(1) and is constant. Similarly, if the input is x1(t) + x2(t), still the output is y(t) = x(1) ≠ x(1) + x(1)

(3.54)

Teaser Verify that the system is non-linear by checking conditions for homogeneity and additivity. (The system is non-linear.) Example 3.12

(Shift invariant, but non-linear system) Check if the following system is shift invariant and linear? y(t) = [x(t)]2. Prove that the system is shift invariant, but non-linear. We will verify the shift invariance property. Let the input be shifted by k time units, therefore the new input is x(t – k). The output is [x(t – k)]2 and is same as shifted by k time units i.e., y(t – k). So, the system is a shift variant system. We will now check for linearity. If the input is kx(t), the output must be k[x(t)]2. However, the actual output is [kx(t)]2 Similarly, if the input is x1(t) + x2(t), the output is not y(t) = [x1(t)+x2(t)]2 ≠ [x1(t)]2 + [x2(t)]2

(3.55)

Teaser Verify that the system is non-linear by checking conditions for homogeneity and additivity. (The system is non-linear.)

CT and DT Systems

Solution

195

Example 3.13

(Linear and shift invariant system) Consider the following system: y(= t ) x(t − 2) Prove that the system is both shift invariant and linear i.e., the system is LTI system. Solution

We will verify the shift invariance property. Let the input be shifted by k units, that is the new input is x(t – k). The output is y(t − k)= x(t − 2 − k).

(3.56)

One can see that the output is also shifted by k units, therefore the system is a shift invariant system. Let us now check the linearity of the system. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

1. Homogeneity: Let the input be doubled. Then the output is also doubled. So, the system is homogeneous. 2. Additivity: We will check the additivity. Let the input be a1x1(t) + a2x2(t) Then the output is a1 y1 (t ) + a2 y2 (= t ) a1 x1 (t − 2) + a2 x2 (t − 2)

(3.57)

Therefore the system is additive. The system satisfies both the homogeneity and additivity properties, therefore the system is linear. It is also shift invariant. Hence, the system is an LTI system. We will solve some examples for DT systems. Example 3.14

(Non-linear and non-shift invariant) Is the following system shift variant? Is it linear? y[n] = x[0] Signals and Systems

Solution

196

We will verify the shift invariance property. Let the input be shifted by k units; therefore, the new input is x[n – k]. The output is still x[0] and is not shifted by k units. So, the system is not a shift variant system Teaser Verify that the system is non-linear. (The system non-linear.) Verify homogeneity and additivity property. Example 3.15

(Shift Invariant however, Non-Linear System) Consider the following system: y[n] = x[n]2 Prove that the system is shift invariant, but non-linear. Solution

We will verify the shift invariance property. Let the input be shifted by k units, that is the new input be x[n – k]. Then the output is given by y[n − k] = x[n − k]2

(3.58)

One can see that the output is also shifted by k units. So the system is a shift invariant system. Let us now check the linearity of the system. Let x[n] = 3, then y[n] = x[n]2 = 9. If the input is now doubled, that is, if x[n] = 6, then the output is y[n] = x[n]2 = 36 ≠ 2 × 9.

(3.59)

We see that the output is not doubled. So, the system is not homogeneous and hence, is not linear. The reader may also check for additivity property. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Example 3.16

(Linear and Shift Invariant System) Consider the following system: y[= n] x[n − 1] Prove that the system is both shift invariant and linear i.e., the system is LTI system. Solution

We will verify the shift invariance property. Let the input be shifted by k units, that is the new input is x[n – k]. The output is y[n − k]= x[n − k − 1].

(3.60)

1. Homogeneity: Let the input be doubled. Then the output is also doubled. So, the system is homogeneous. 2. Additivity: We will check the additivity. Let the input be a1x1[n] + a2x2[n] Then the output is a1 y1[n] + a2 y2 [= n] a1 x1[n − 1] + a2 x2 [n − 1]

(3.61)

Therefore, the system is additive. Since the system satisfies both the homogeneity and additivity properties, therefore the system is linear. It is also shift invariant. Hence, the system is LTI system. We will now prove the property of shift invariance for following CT systems.

CT and DT Systems

One can see that the output is also shifted by k units, therefore the system is a shift invariant system. Let us now check the linearity of the system.

197

Example 3.17

Consider the CT systems given by 1.

y(t ) = cos(x(t ))

2.

y(t ) = ∫ x(τ )dτ

3.

y(t ) =

t /4

−∞

d x(t ) dt

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4.

y(= t ) x(3 − t )

5.

y(t ) = x(t / 5)

Are the systems shift/time invariant? If not, explain why. Solution

1. Consider a system given by y(t) = cos(x(t)) Let the input be shifted by k units that is the new input is x(t – t). The output is y(t −= τ ) cos(x(t − τ ))

(3.62)

One can see that the output is also shifted by τ units, therefore the system is a shift invariant system. 2.

t /4

−∞

x(τ )dτ Let the input be shifted by k units that is the new input is

x(t – t). The output is

Signals and Systems

198

y(t ) = ∫

(t −τ )/4

y(t − τ ) = ∫ −

x(τ )dτ .

(3.63)

One can see that the output is also shifted by τ units, therefore the system is a shift invariant system. 3.

d x(t ) Let us verify the superposition property. dt Let the input be shifted by k units that is the new input is x(t – t). The output is y(t ) =

y(t −= τ)

d (x(t − τ )). dt

(3.64)

One can see that the output is also shifted by τ units, therefore the system is a shift invariant system. 4. y(t) = x(3 – t) Let the input be shifted by k units that is the new input is x(t – t). The output is y(t − τ )= x(3 − (t − τ )).

(3.65)

One can see that the output is also shifted by τ units, therefore the system is a shift invariant system. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

5. y(t) = x(t/5) Let the input be shifted by k units that is the new input is x(t – t). The output is y(t − τ )= x((t − τ ) / 5).

(3.66)

One can see that the output is also shifted by τ units, therefore the system is a shift invariant system. Example 3.18

1.

y[n] = sin(x[n])

2.

y[n] = 3x[n]u[n]

3.

y[n] = log10(|x[n]|)

y[n] 4. =

CT and DT Systems

Consider the DT systems given by

n

∑ x[k + 1]

k = −∞

∞

= 5. y[n] x[n] ∑ δ [n − 3k] k = −∞

6.

y[n] = x[n2]

199

Are the systems shift/time invariant? If not, explain why. Solution

1. y[n] = sin(x[n]) We will verify the shift invariance property. Let the input be shifted by k units, that is the new input is x[n – k]. The output is y[n −= k] sin(x[n − k])

(3.67)

One can see that the output is also shifted by k units, therefore the system is a shift invariant system. 2. y[n] = 3x[n]u[n] We will verify the shift invariance property. Let the input be shifted by k units, that is the new input is x[n – k]. The output is y[n − k]= 3x[n − k]u[n] ≠ 3x[n − k]u[n − k]

(3.68)

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One can see that the output is not shifted by k units, therefore the system is not a shift invariant system. 3. y[n] = log10(|x[n]|) We will verify the shift invariance property. Let the input be shifted by k units, that is the new input is x[n – k]. The output is y[= n − k] log10 (| x[n − k]|)

(3.69)

One can see that the output is also shifted by k units, therefore the system is a shift invariant system. y[n] 4. =

n

∑ x[k + 1].

k = −∞

Signals and Systems

We will verify the shift invariance property. Let the input be shifted by k units, that is the new input is x[n – k]. The output is

200

y[n −= k]

n−k

∑ x[k + 1]

(3.70)

k = −∞

One can see that the output is also shifted by k units, therefore the system is a shift invariant system. ∞

= 5. y[n] x[n] ∑ δ [n − 3k] k = −∞

We will verify the shift invariance property. Let the input be shifted by k1 units, that is the new input is x[n – k1]. The output is ∞

∞

k = −∞

k = −∞

y[n − k1 ] = x[n − k1 ] ∑ δ [n − 3k] ≠ x[n − k1 ] ∑ δ [n − 3k − k1 ]

(3.71)

One can see that the output is not shifted by k units, therefore the system is not a shift invariant system. 6. y[n] = x[n2] We will verify the shift invariance property. Let the input be shifted by k units, that is the new input is x[n – k]. The output is y[n − k]= x[(n − k )2 ]

(3.72)

One can see that the output is also shifted by k units, therefore the system is a shift invariant system.

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Concept Check

• • • • • • • • •

What is meaning of a linear system? What is the property of homogeneity? What is additivity property? What is the physical significance of linearity? When will you call the system as incrementally linear? Can a system with linear graph be non-linear? Can a system with non-linear transfer curve be linear? What is the physical significance of shift invariance? What is LTI system?

3.2 Properties of CT and DT Systems – Causality and Memory

3.2.1

Causality property A system is said to be causal if its present output depends only on past and present inputs and past outputs. Let us consider an example of a causal CT and DT system. y(= t ) a0 x(t ) + a1 x(t − 1) − b1 y(t − 1)

(3.73)

y[= n] a0 x[n] + a1 x[n − 1] − b1 y[n − 1]

(3.74)

CT and DT Systems

We will study the properties of causality and memory in this section.

201

Here, the output at time instant ‘t’ in case of CT system or ‘n’ in case of DT system depends on current input at time instant ‘t’ or ‘n’ and past input and output at instant ‘t – 1’ or ‘n – 1’ i.e., previous time sample. These systems are causal. Now consider CT and DT systems represented as y(= t ) a0 x(t ) + a1 x(t + 1) − b1 y(t − 1)

(3.75)

y[= n] a0 x[n] + a1 x[n + 1] − b1 y[n − 1]

(3.76)

These are non-causal systems as the output of the system at current time instant ‘t ’ or ‘n’ depends on the current inputs and the next input at time instant ‘t + 1’ and ‘n + 1’, respectively.

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Physical significance of Causality Causal systems are practically realizable or implementable. The system of a human being is causal as we always keep on learning from the past inputs and past outputs of the system. The future inputs have no effect on our act at current or present time. The causal systems can be implemented in real time. The present and past inputs have meaning only for temporal systems where time is an independent variable. In case of spatial domain systems, present and past input has no meaning. Such noncausal systems can be implemented after grabbing the complete input data. Non-causal temporal systems can be implemented if some delay is tolerable. The system considers the future inputs for calculating the current outputs and can generate a bench mark for system performance. The real-time system performance may be compared with the ideal system implemented with some tolerable delay. Offline systems can always be implemented as non-causal systems. We will consider some examples for CT systems first. Example 3.19 Signals and Systems

Consider a system represented by

202

= y(t )

τ =1

∫τ

= −1

h(τ )x(t − τ )dτ

Is the system causal? If not, explain why. Solution

In this case, h(t) exists for τ = –1. The input for this value of time is t + 1. The output of the system at time instant t depends on the next input, i.e., input at t + 1, therefore the system is non-causal. Example 3.20

Consider a system given by y(t) = x(t – 1) + y(t – 3) Is the system causal? Solution

The output of the system at instant t depends on input at current time instant t and past input at t = t – 1 and past output at time instant t = t – 3, therefore the system is causal. Example 3.21

Consider a system represented as y(t) = x(1 – t). Is it causal? Solution

The system output at time instant t depends on time instant 1 – t. consider t = –2, 1 – t becomes 1 + 2 i.e., 3. So, the output at t = –2 time units depends on input at time instant t = 3. This is a future time instant. Hence, the system is non-causal. We will now consider examples for DT systems. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Example 3.22

Consider a system represented by = y[n]

1

∑ h[k]x[n − k]

k = −1

Is the system causal? If not, explain why. Solution

In this case, h[n] exists for n = –1. From the given equation we have y[n] = h[−1]x[n + 1] + h[0]x[n] + h[1]x[n − 1]

(3.77)

The output of the system at instant n depends on the next input, i.e., input at n + 1, therefore the system is non-causal. Example 3.23

= y[n]

CT and DT Systems

Consider the system represented by n +1

∑ x[k]h[n − k]

k = −1

Is the system causal? If not, explain why. Solution

203

Expanding the given equation we have y[n] = x[−1]h[n + 1] + x[0]h[n] + ...... + x[n + 1]h[−1]

(3.78)

We see from Eq. (3.78) that the output of the system at n depends on the input at the next time instant, that is, at n + 1, if h[n] exists for n = –1 and therefore the system becomes non-causal. Example 3.24

Consider a system given by y[n]= x[n] − x[n − 1] + y[n − 2] Is the system causal? Solution

The output of the system at instant n depends on current and past input and past output; therefore, the system is causal.

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Example 3.25

Consider a system given by y[n]= x[−n] Is the system causal? If not, explain why. Solution

The system is non-causal. Reason will get

To see why it is non-causal, put n = –2 in the given equation. We

y[–2] = x[2]

(3.79)

The output at n = –2 depends on the input at the next time instant, that is, n = 2; therefore, the system is non-causal; Signals and Systems

Example 3.26

Consider the CT systems given by 1.

y(t) = cos(x(t))

2.

y(t ) = ∫

3.

y(t ) =

4.

y(t) = x(3 – t)

5.

y(t) = x(t/5)

t /4

−∞

204

x(τ )dτ

d x(t ) dt

Are the systems causal? If not, explain why. Solution

1. y(t) = cos(x(t)). Here, the system output at current instant of time ‘t’ depends on only the current input at time ‘t’. Hence, the system is causal. t /4

2. y(t ) = ∫−∞ x(τ )dτ . Here, the system output is the integration of its previous inputs from minus infinity to t/4. Let us put t = 4, now we have to integrate the inputs up to t = 1. Let us put t = –4, now we have to integrate the inputs up to t = –1, which is the next time instant. Hence, the system is noncausal. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

d x(t ). Here, the system output at current instant of time instant ‘t’ dt depends on only the current input at ‘t’. Hence, the system is causal. 4. y(t) = x(3 – t). Let us put t = –1. The output at time instant of –1 depends on input at 4 = 3 – (–1) i.e., the next time instant. Hence, the system is non-causal. 5. y(t) = x(t/5). Let us put t = –5, the output at t = –5 depends on input at t = –1, which is the next time instant. Hence, the system is non-causal. 3.

y(t ) =

Example 3.27

1.

y[n] = sin(x[n])

2.

y[n] = 3x[n]u[n]

3.

y[n] = log10(|x[n]|)

y[n] 4. =

n

∑ x[k + 1]

k = −∞

∞

= 5. y[n] x[n] ∑ δ [n − 3k] k = −∞

6.

y[n] = x[n2]

CT and DT Systems

Consider the DT systems given by

205

Are the systems causal? If not, explain why. Solution

1. y[n] = sin(x[n]). Here, the system output at current instant of time ‘n’ depends on only the current input at ‘n’. Hence, the system is causal. 2. y[n] = 3x[n]u[n]. Let us first write the definition of u[n]. u[n] = 1 for n ≥ 0 = 0 otherwise Let us put n = –1, u[–1] = 0, The output at time ‘–1’ is zero irrespective of the input x[n]. Let us put n = +1, u[+1] = 1, The output at time ‘1’ is 3x[1] depend on current input x[1]. The system is causal. 3. y[n] = log10(|x[n]|). Let us put n = –1, the output at t = –1 will depend on log of x[–1], i.e., current time instant input. The system is causal. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

y[n] 4.=

n

∑ x[k + 1]. Let us put n = 1, the output at instant 1 depends on

k = −∞

inputs from minus infinity to n = 2 i.e., the next time instant. The system is non-causal. ∞

= 5. y[n] x[n] ∑ δ [n − 3k]. The output depends only on the current sample; if k = −∞

the delta function exists, then the system is causal. 6. y[n] = x[n2]. Consider n = 1, the output at time instant 1 depends on input sample at time instant 1; however, the output at time instant 2 depends on the input at time instant 4. This makes the system non-causal.

Signals and Systems

3.2.2 Memory

206

The system is said to have memory or said to be dynamic if its current output depends on previous, future input or previous and future output signals. The system is said to be memoryless or instantaneous if its current output depends only on the current input. The extent of memory depends on the number of past inputs or outputs on which the current output depends. We will consider examples of systems with memory. Example 3.28

Consider a CT system given by t

y(t ) = ∫ x(τ )dτ −∞

Does the system have memory? Solution

The current output of the system depends on past inputs. The system is with memory. The memory extends into infinite past inputs. Example 3.29

Consider a CT system given by y(t ) = 4 x(t ) Does the system have memory? Solution

The current output of the system depends only on the current input. The system is without memory or is memoryless.

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Example 3.30

Consider a DT system given by y[n=]

1 [x[n] + x[n − 1]] 2

Does the system have memory? Solution

The current output of the system depends on current and one past input. The system is with memory. The memory extends into one past input. Example 3.31

Consider a DT system given by y[n]= 5 + 4 x[n]

Solution

The current output of the system depends only on the current input. The system is without memory or is memoryless. Physical significance of memory The systems containing only the resistive network has no memory and is said to be memoryless or instantaneous systems. If the system contains active element like capacitor or inductor, the system memory will extend up to infinity. Such systems are said to be dynamic. Generally, we have to handle dynamic systems. We human beings always learn from the past. Our brain system has a memory that allows us to understand the new concepts.

CT and DT Systems

Does the system have memory?

207

Example 3.32

Consider the CT systems given by 1.

y(t) = cos(x(t))

2.

y(t ) = ∫

3.

y(t ) =

4.

y(t) = x(3 – t)

5.

y(t) = x(t/5)

t /4

−∞

x(τ )dτ

d x(t ) dt

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Do the systems have memory? If yes, find the extent of memory. Solution

1. y(t) = cos(x(t)). Here, the system output at current instant of time ‘t’ depends on only the current input at ‘t’. Hence, the system is memoryless. t /4

Signals and Systems

2. y(t ) = ∫ x(τ )dτ . Here, the system output is the integration of its previous −∞ inputs from minus infinity to t/4. Hence, the system has memory and it extends from infinity to t/4. d 3. y(t ) = x(t ). Here, the system output at current instant of time instant ‘t’ dt depends on only the current input at ‘t’. Hence, the system is memoryless.

208

4. y(t) = x(3 – t). Let us put t = 2. The output at time instant of 2 depends on input at t = 3 –(2) = 1 i.e., the previous time instant. Hence, the system has memory. The extent of memory varies with the value of the current instant. 5. y(t) = x(t / 5). The output at the current instant depends on the input at previous instant t/5. The system has memory. The extent of memory is up to t/5 time. Example 3.33

Consider the DT systems given by 1.

y[n] = sin(x[n])

2.

y[n] = 3x[n]u[n]

3.

y[n] = log10(|x[n]|)

y[n] 4. =

n

∑ x[k + 1]

k = −∞

∞

= 5. y[n] x[n] ∑ δ [n − 3k] k = −∞

6.

y[n] = x[n2]

Do the systems have memory? If yes, find the extent of memory. Solution: 1. y(t) = sin(x[n]). Here, the system output at current instant of time ‘n’ depends on only the current input at ‘n’. Hence, the system is memoryless. 2. y[n] = 3x[n]u[n]. Here, the system output at current instant of time ‘n’ depends on only the current input at ‘n’. Hence, the system is memoryless.

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3. y[n] = log10(|x[n]|). Here, the system output at current instant of time ‘n’ depends on only the current input at ‘n’. Hence, the system is memoryless. 4.= y[n]

n

∑ x[k + 1]. Here, the system output at current instant of time ‘n’

k = −∞

depends on one future input, current input and past inputs form infinity to previous input. Hence, the system has memory. ∞

5. y[n] x[n] ∑ δ [n − 3k]. Here, the system output at current instant of time = k = −∞

‘n’ depends on current input if the delta function exists. Hence, the system is memoryless. 6. y[n] = x[n2]. Consider n = 1, the output at time instant 1 depends on input sample at time instant 1; however, the output at time instant 2 depends on the input at time instant 4. The system has memory; however, it depends on future inputs and the extent of memory depends on the value of ‘n’ current sample. Concept Check

Explain the meaning of causality.

Can you realize a non-causal system?

CT and DT Systems

• • • • • •

What are real-time systems?

209

What is physical significance of causality? When will you call the system as memoryless? What is the physical significance of system with memory?

3.3 Properties of CT and DT Systems – Invertibility and Stability We will study the properties of invertibility and stability in this section.

3.3.1 Invertibility If it is possible to recover the input of the system, then the system is said to be invertible. Consider the system represented by the operator H producing the output as y(t) with input as x(t). The idea of system invertbility is more clearly understood if one considers the new operator H–1called as inverse operator with the so-called inverse system, as shown in Fig. 3.2. The H–1 is not the reciprocal of the operator H; however, it is the symbol used to indicate the inverse.

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Fig. 3.2 Invertibility of systems

Signals and Systems

Note that there must be a one-to-one correspondence between the input and the output in order to have invertibility.

210

Physical significance of invertibility When we use a system for processing the signal, for example, for calculating some transform for removing the noise, the system is required to be invertible (inverse transform) to recover the signal back in time domain. If we process the signal to extract some features for recognition, then the system may be non-invertible as we are not recovering the signal. Invertible system also finds applications in communication field. For error-free transmission an equalizer is used at the input of the receiver that has inverse characteristics as that of the channel. We will consider some examples of invertible systems. Example 3.34

Consider a CT system given by t

y(t ) = ∫ x(τ )dτ −∞

Is the system invertible? Solution

The output of the system can be differentiated to get the signal x(t). Differentiator is the inverse system and the integrator is said to be the invertible system. We can obtain x(t) by differentiating the output y(t) d y(t ) = x(t ) dt

(3.80)

Example 3.35

Consider a CT system given by ∞

y( jω ) = ∫ x(τ )e − jωτ dτ −∞

Is the system invertible?

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Solution

The output of the system can be recognized as the FT of the input signal x(t). Inverse FT is the inverse system and FT is said to invertible system. We can obtain x(t) by taking Inverse FT of the output y(t) ∞

x(t ) = ∫ y( jω )e jωτ dω −∞

(3.81)

Example 3.36

Consider a DT system given by = y[n] 3x[n] + 4 Is the system invertible? Solution

The input can be recovered from the output using the equation (3.82)

CT and DT Systems

1 [ y[n] − 4] 3

Example 3.37

Consider a DT system given by

211

= x[n]

The system is invertible.

N −1

X[K ] = ∑ x[n]e − j 2π nk / N n=0

Is the system invertible? Solution

The output of the system can be recognized as the DFT of the input signal x[n]. Inverse DFT is the inverse system and DFT is said to invertible system. We can obtain x[n] by taking inverse DFT of the output x[k]. N −1

x[n] = ∑ X[k]e j 2π nk / N

(3.83)

k =0

Example 3.38

Consider a CT system given by y(t ) = (x(t ))2 Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Is the system invertible? Solution

The output of the system can be recognized as the square of the input signal x(t). We can obtain x(t) by taking the square root of the output. x(t ) = ± y(t )

(3.84)

There is no one-to-one correspondence. Hence, the system is not invertible. Example 3.39

Signals and Systems

Consider the CT systems given by

212

1.

y(t) = cos(x(t))

2.

y(t ) = ∫

3.

y(t ) = ∫ x(τ )dτ

4.

y(t ) =

5.

y(t) = x(3 – t)

6.

y(t) = x(t/5)

t /4

−∞

x(τ )dτ

t

t −3

d x(t ) dt

Are the systems invertible? If not, find why they are not invertible. Solution

1.

y(t) = cos(x(t))

= x(t ) cos −1 ( y(t )) ± 2nπ

(3.85)

There is no one-to-one correspondence between x(t) and y(t). Hence, the system is non-invertible. 2. y(t ) = ∫

t /4

−∞

x(t / 4) =

x(τ )dτ

d y(t ) dt

(3.86)

The system is invertible. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

3.

t

y(t ) = ∫ x(τ )dτ t −3

= y(t )

∫

t

−∞

x(τ )dτ − ∫

t −3

−∞

x(τ )dτ

d x(t ) − x(t − 2) = y(t ) dt

(3.87)

The system is invertible. 4.

y(t ) =

d x(t ) dt t

x(t ) = ∫ y(τ )dτ −∞

(3.88)

There is one-to-one correspondence. The system is invertible. y(t) = x(3 – t) x(= t ) y(t + 3)

(3.89)

The system is invertible. 6.

y(t) = x(t/5) x(t ) = y(5t )

(3.90)

CT and DT Systems

5.

213

The system is invertible. Example 3.40

Consider the DT systems given by 1.

y[n] = sin(x[n])

2.

y[n] = 3x[n] – 5

3.

y[n] = 3x[n]u[n]

4.

y[n] = log10(|x[n]|)

y[n] 5. = 6.

n

∑ x[k + 1]

k = −∞

y[n] = x[n2]

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Are the systems invertible? If yes, find the reason for invertibility. Solution

1. y[n] = sin(x[n]) = x[n] sin −1 ( y[n]) ±

1 2

(3.91)

There is no one-to-one correspondence. The system is non-invertible. 2. y[n] = 3x[n] – 5 = x[n]

1 ( y[n] + 5) 3

(3.92)

The system is invertible. 3. y[n] = 3x[n]u[n]

Signals and Systems

∞

214

y[n] = ∑ 3x[m] m=0

= x[n]

∞ 1 y[n] − ∑ 3x[m] 3 m =1

(3.93)

The system is invertible.

4. y[n] = log10(|x[n]|) (3.94)

x[n] = ±10 y[n]|

The system is non-invertible. There is no one-to-one correspondence. y[n] 5. =

n

∑ x[k + 1]

k = −∞

x[n]= y[n] − x[n + 1] −

n−2

∑ x[k + 1]

k = −∞

(3.95)

Here, the system is invertible. 6. y[n] = x2[n] x[n] = ± y[n]

(3.96)

The system is non-invertible, as there is no one-to-one correspondence.

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3.3.2 Stability Stability is a notion that describes whether the system will be able to follow the input. A system is said to be unstable if its output is out of control or increases without bound. To define stability, we define the following types of responses for linear time invariant systems. 1. Zero State Response This is due to the input only. All initial conditions are zero. 2. Zero Input Response This is due to the initial conditions only. All inputs are zero. The system response will be discussed in detail in chapter 4. BIBO stability An arbitrary relaxed system (with zero initial conditions) is said to be bounded input bounded output (BIBO) stable if and only if its output is bounded for every bounded input.

if | x[n]|< ∞ then | y[n]|< ∞ for − ∞ ≤ n ≤ ∞

(3.97)

For zero input stability, the magnitude of the output signal/sequence must be finite and it must asymptotically approach zero. If the output sequence is represented as y[n], then | y(t )|< ∞; limt →∞ | y(t )|= 0 | y[n]|< ∞; limn→∞ | y[n]|= 0

CT and DT Systems

if | x(t ) |< ∞ then | y(t ) |< ∞ for − ∞ ≤ t ≤ ∞

215

(3.98)

Physical significance of Stability The stable system is only of practical value. If the system is unstable, the extraneous noise or even thermal noise may allow the system to become unstable and the system output may become infinite. The characteristic roots of the system will basically decide the stability of the system. (Refer to section 4.2.) We will consider some examples to illustrate the concepts. Example 3.41

Consider a system given by y(t) = x(t) + 3. Determine if the system is stable. Solution

Let the input signal be bounded. | x(t )|< M for − ∞ ≤ t ≤ ∞ The output can be written as | y(= t )| | x(t )| +3 < M + 3

(3.99)

The output of the system is also bounded. Hence, the system is stable. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Example 3.42 t

Consider a system given by y(t ) = ∫ x(τ )dτ −∞

Solution

Let the input be bounded. | x(t )|< M for − ∞ ≤ t ≤ ∞ The output integrates the values of input from infinity to t. Hence, infinite values are used in the integration, which may lead to infinite output. The system cannot be guaranteed to be stable. Example 3.43 10

Consider a system given by y[n] = ∑ x[m] m=0

Solution

Signals and Systems

Let the input be bounded. | x(t )|< M for − ∞ ≤ t ≤ ∞ . The output sums the values of input from m = 0 to m = 10. Hence, only 11 values are used in the summation that results in the output of

216

(3.100)

| y(t )|< 11M for − ∞ ≤ t ≤ ∞ The system output is bounded and hence, the system is stable. Example 3.44

Consider a system given by y[n] =

n

∑ x[m]

m = −∞

Solution

Let the input be bounded. The output sums the values of input from infinity to n. Hence, infinite values are used in the summation that may lead to infinite output. The system cannot be guaranteed to be stable. Example 3.45

Consider the CT systems given by 1.

y(t) = cos(x(t))

2.

(t ) = ∫

3.

y(t ) = ∫ x(τ )dτ

4.

y(t ) =

t /4

−∞

x(τ )dτ

t

t −3

d x(t ) dt

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5.

y(t) = x(3 – t)

6.

y(t) = x(t/5)

Are the systems stable? If not, find out why they are not stable. Solution

1. y(t) = cos(x(t)) Let the input be bounded. | x(t ) |< M for − ∞ ≤ t ≤ = ∞ | y(t ) | | cos(x(t )) |≤ 1

(3.101)

The cosine terms are always bounded and have a value less than 1. Hence, the system is stable. y(t ) = ∫

t /4

−∞

x(τ )dτ Let the input be bounded.

| x(t ) |< M for − ∞ ≤ t ≤ ∞ | y(t ) |=| ∫

t /4

−∞

x(τ )dτ |

(3.102)

The output is the integration of values between minus infinity to t/4 i.e., infinite values. The output cannot be guaranteed to be bounded. Hence, the system is unstable. 3.

t

y(t ) = ∫ x(τ )dτ t −3

Let the input be bounded.

CT and DT Systems

2.

217 t

| x(t ) |< M for − ∞ ≤ t ≤ ∞, | y(t ) |=| ∫ x(τ )dτ | t −3

(3.103)

The output is the integration of values between t – 3 to t i.e., a finite time interval of 3 time units. Hence, the system is stable. 4.

y(t ) =

d x(t ) dt

Let the input be bounded. | x(t ) |< M for − ∞ ≤ t ≤ ∞, | y(t ) |=|

d | x(t ) || dt

(3.104)

The output is the differentiation of a finite quantity. Hence, the system is stable. 5. y(t) = x(3 –t) Let the input be bounded. | x(t ) |< M for − ∞ ≤ t ≤ ∞, | y(= t ) | | x(3 − t ) |< M

(3.105)

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Hence, the system is stable. 6. y(t) = x(t/5) Let the input be bounded. | x(t ) |< M for − ∞ ≤ t ≤ ∞, = | y(t ) | | x(t / 5) |< M

(3.106)

Hence, the system is stable. Example 3.46

Signals and Systems

Consider the DT systems given by

218

1.

y[n] = sin(x[n])

2.

y[n] = 3x[n] –5

3.

y[n] = 3x[n]u[n]

4.

y[n] = log10(|x[n]|)

y[n] 5. =

6.

n

∑ x[k + 1]

k = −∞

y[n] = x[n2]

Are the systems stable? If yes, find the reason for stability. Solution

1. y[n] = sin(x[n]) Let the input be bounded. | x[n]|< M for − ∞ ≤ n ≤ ∞ | y[n]| = | sin x([n]) |≤ 1

(3.107)

The sine terms are always bounded and have a value less than 1. Hence, the system is stable. 2. y[n] = 3x[n]–5 Let the input be bounded. | x[n]|< M for − ∞ ≤ n ≤ ∞, | y[n]| = | 3 | x[n]|| −5 ≤ 3 M − 5

(3.108)

The system output is bounded. Hence, the system is stable. 3. y[n] = 3x[n]u[n] Let the input be bounded.

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| x[n]|< M for − ∞ ≤ n ≤ ∞, | y[n]| = | 3 | x[n]u[n]||< M

(3.109)

The system output is bounded. Hence, the system is stable. 4. y[n] = log10(|x[n]|) Let the input be bounded. | x[n]|< M for − ∞ ≤ n ≤ ∞, | y[n]| = | log10 | x[n]||< M

(3.110)

The system output is bounded. Hence, the system is stable. 5. = y[n]

n

∑ x[k + 1]

k = −∞

Let the input be bounded. n

∑ x[k + 1]|→ ∞

k = −∞

(3.111)

The system output is the summation of infinite values. Hence, the system is unstable. 6. y[n] = x2[n] Let the input be bounded. | x[n]|< M for − ∞ ≤ n ≤ ∞, | = y[n]| | x 2 [n]|< M 2

(3.112)

The system output is the square of a finite value. Hence, the system is stable.

CT and DT Systems

| x[n]|< M for − ∞ ≤ n ≤ ∞, | y[n]| = |

219

Concept Check

• • • • • • •

What is meaning of invertibility? Specify a situation when you need the invertibility property. What is the physical significance of invertibility? How will you confirm the stability of the system? What will happen to the output of the system if the system is unstable? What is the essential condition for stability of a system? What is BIBO stability?

3.4 System Representation as Interconnection of Operations This section interprets the systems as interconnection of operations. Let us consider a simple example to illustrate the concept. When the input signal is Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

transformed by some operator H to produce the output signal, the operator denotes the action of the system. Let the output of any system y(t) or y[n] be represented as y(t) = H[x(t)] y[n] = H[x[n]]

(3.113)

Where, H is the operator that acts on the input signal x(t) or x[n] to produce the output y(t) or y[n]. The operator H can be represented as interconnection of operators as follows. Consider a system represented by the Eq. (3.114). ) y(t=

Signals and Systems

y[n=]

220

1 [ x(t ) + x(t − 1) + x(t − 2)] 3

(3.114)

1 [ x[n] + x[n − 1] + x[n − 2]] 3

This equation represents a moving average system. Basically, the current output of the system is found by taking the average of current and previous two input samples. To obtain the next output sample, we have to take the average of next, current and the previous input sample. This is a sliding average and hence, called as the moving average system. Let us try to represent the system in the form of a block schematic, as shown in Fig. 3.1. Let us represent the time shifting operation by the operator S. Here, S represents a time shift by 1 time unit for CT systems and time shift by 1 sample for DT systems. Now, the operator H can be represented as H= y(t )=

1 [1 + S + S 2 ] 3

and

1 [1 + S + S 2 ]x(t ) 3

1 y[n]= [1 + S + S 2 ]x[n] 3

(3.115)

The representation is clear from Fig. 3.3 drawn for a DT system. A similar representation will hold good for CT system as well. We will go through further examples to illustrate the concept.

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Fig. 3.3 Block schematic for a system

Consider a CT system represented by Eq. (3.116), represent it as the interconnection of operators. 1 1 1 y(t ) =[ x(t ) + x(t − 2) + x(t − 4) + x(t − 6)] 2 3 2

(3.116)

CT and DT Systems

Example 3.47

221

Solution

Let us try to represent the system in the form of a block schematic, as shown in Fig. 3.4. Let us represent the time shifting operation by the operator S, which represents a time shift by 2 time units. Now, the operator H can be represented as 1 2 1 3  1 1 1   1 H =1 + S + S 2 + S 3  and y(t ) =1 + S + S + S  x(t ) 3 2  3 2   2  2

(3.117)

The representation is clear from Fig. 3.4. Example 3.48

Consider a CT system represented by the Eq. (3.118). 1 1   y(t ) =  x(t ) + x(t − 1) − y(t − 2) 2 3  

(3.118)

Represent it as the interconnection of operators. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Fig. 3.4 Block schematic for Example 3.47

Signals and Systems

Solution

222

Let us try to represent the system in the form of a block schematic, as shown in Fig. 3.5. Let us represent the time shifting operation by the operator S that represents a time shift by 1 time unit. Now, the operator H can be represented as  1   1 2  1   1 2 H= 1 + 2 S  / 1 + 3 S  x(t ) 1 + 2 S  / 1 + 3 S  and y(t ) =        

(3.119)

Note that we have to take the term containing y(t – 2) on the left-hand side to combine it with y(t). The representation is clear from Fig. 3.5.

Fig. 3.5 Block schematic for Example 3.48 Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Example 3.49

Consider a DT system represented by Eq. (3.120), represent it as the interconnection of operators. 1 1 1 y[n] =[ x[n] + x[n − 1] + x[n − 2] + x[n − 3]] 2 3 2

(3.120)

Solution

Let us try to represent the system in the form of a block schematic, as shown in Fig. 3.6. Let us represent the time shifting operation by the operator S. Now, the operator H can be represented as 1 2 1 3  1 1 1   1 H =1 + S + S 2 + S 3  and y[n] =1 + S + S + S  x[n] 2 3 2  2 3 2   

(3.121)

Example 3.50

Consider a system represented by the Eq. (3.122). 1 1 y[n] =[ x[n] + x[n − 1] − y[n − 1]] 2 3

(3.122)

CT and DT Systems

The representation is clear from Fig. 3.6.

223

Fig. 3.6 Block schematic for Example 3.49 Represent it as the interconnection of operators.

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Solution

Let us try to represent the system in the form of a block schematic, as shown in Fig. 3.7. Let us represent the time-shifting operation by the operator S. Now, the operator H can be represented as  1   1  and  1   1  H= y[n] = 1 + 2 S  / 1 + 3 S  1 + 2 S  / 1 + 3 S  x[n]        

(3.123)

Signals and Systems

Note that we have to take the term containing y(n – 1) on the left-hand side to combine it with y(n). The representation is clear from Fig. 3.7.

224

Fig. 3.7 Block schematic for Example 3.50 Concept Check

• What is a moving average system? • How will you divide the system operator into number of operators such as time sifting operator?

• What is the advantage of using interconnection of operations?

3.5 Series and Parallel Interconnection of Systems Let us understand the significance of series and parallel interconnections of the systems. We need to understand what will happen to the impulse response of the interconnected system. These interconnections have significance for LTI systems.

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3.5.1

Series interconnection of systems Consider the series connection of two linear and shift invariant systems, i.e., LTI CT systems, as shown in Fig. 3.8. Let the impulse responses of the two systems be specified as h1(t) and h2(t), respectively. The series connection is also termed as cascaded configuration. The output of the system is y(t). Let z(t) represent the output of the first system with impulse response h1(t). The output z(t) can be written as z (τ ) = x(τ ) * h1 (τ )

∫

∞

−∞

(3.124) x(η )h1 (τ − η )dη

Here, the symbol * denotes the convolution operation. The meaning of convolution and the procedure to implement convolution in CT and DT domain will be discussed in detail in chapter 4. The output of first system z(t) is applied as input to the second system with impulse response h2(t). The output of the second system y(t) will be given as y(t ) = z (t ) * h2 (t ) = y(t ) = y(t )

∫

∞

−∞

z (τ )h2 (t − τ )dτ

∞

∫ ∫

∞

−∞ −∞

CT and DT Systems

= z (τ )

225

x(η )h1 (τ − η )h2 (t − τ )dηdτ

υ, put τ − η = y(t ) = ∫

∞

∫

∞

−∞ −∞

x(η )h1 (υ )h2 (t − υ − η )dηdυ

∞

∞

−∞

−∞

(3.125)

y(t ) = ∫ x(η )∫ h1 (υ )h2 (t − η − υ )dυ = y(t )

∫

∞

−∞

x(η )h(t − η )dη

h(t ) = h1 (t ) * h2 (t ) y(t ) = = x(t ) * h(t )

∫

∞

−∞

x(η )h(t − η )dη

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We can conclude that when the two systems are cascaded or connected in series, the impulse response of the cascaded system is the convolution of impulse responses of the two systems.

Signals and Systems

Fig. 3.8 Series interconnection of CT systems In case of DT systems, same equations will hold good. When the two systems with impulse responses h1[n] and h2[n] are cascaded or connected in series, the impulse response of the cascaded system is the convolution of impulse responses of the two systems given by h1[n]*h2[n]. Consider the series connection of two linear and shift invariant systems, i.e., LTI DT systems, as shown in Fig. 3.9. Let the impulse responses of the two systems be specified as h1[n] and h2[n], respectively. The series connection is also termed as cascaded configuration. The output of the system is y[n]. Let z[n] represent the output of the first system with impulse response h1[n]. The output z[n] can be written as z[n] = x[n]* h1[n]

226 z[n] =

(3.126)

n

∑ x[m]h [n − m] 1

m = −∞

The output of first system z[n] is applied as input to the second system with impulse response h2[n]. The output of the second system y[n] will be given as y[n] = z[n]* h2 [n] y[n] =

n

∑ z[m]h [n − m] 2

m = −∞

= y[n]

n

m

∑ ∑ x[l]h [m − l]h [n − m]

m = −∞ l = −∞

1

2

put m – l = w

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y[n] =

n −l

m

∑ ∑ x[l]h [w]h [n − l − w]

w = − l −∞ l = −∞

= y[n]

1

2

(3.127)

n

∑ x[l]h[n − l]

l = −∞

h[n] = h1[n]* h2 [n] y[n] = x[n]* h[n]

Fig. 3.9 Series interconnection of DT systems

3.5.2 Parallel interconnection of systems Consider the parallel connection of two linear and shift invariant systems, i.e., LTI systems, as shown in Fig. 3.10. Let the impulse responses of the two systems be specified as h1(t) and h2(t), respectively.

CT and DT Systems

We can conclude that when the two systems are cascaded or connected in series, the impulse response of the cascaded system is the convolution of impulse responses of the two systems.

227

Fig. 3.10 Parallel interconnection of CT systems The output of the system is y(t). Let z(t) represent the output of the first system with impulse response h1(t). Let w(t) represent the output of the second system with impulse response h2(t). The output z(t) can be written as Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

z(t) = x(t) *h1(t) = z (t )

∞

∫

−∞

x(η )h1 (t − η )dη

(3.128)

The symbol * is used to represent the convolution operation. The output w(t) can be written as w(t) = x(t) *h2(t) = w(t )

∫

∞

−∞

x(η )h2 (t − η )dη

(3.129)

The output of the parallel interconnection y(t) can be written as

Signals and Systems

= y(t ) x(t ) * h1 (t ) + x(t ) * h2 (t )

228

= y(t )

∫

∞

= y(t )

∫

∞

−∞

−∞

∞

x(η )h1 (t − η )dη + ∫ x(η )h2 (t − η )dη −∞

(3.130)

x(η )[h1 (t − η ) + h2 (t − η )]dη

Let the impulse response of the parallel configuration be written as h(t) given by h(t) = h1 (t) + h2(t) = y(t )

∫

∞

y= (t )

∫

∞

−∞

−∞

x(η )[h1 (t − η ) + h2 (t − η )]dη

(3.131)

x(η )h(t − η= )dη x(t ) * h(t )

We can conclude that when the two systems are connected in parallel, the impulse response of the parallel configuration system is the addition of impulse responses of the two systems. In case of DT systems, same equations will hold good. When the two systems with impulse responses h1[n] and h2[n] are connected in parallel, the impulse response of the cascaded system is the addition of impulse responses of the two systems given by h1[n] + h2[n]. Consider the parallel connection of two linear and shift invariant systems, i.e., LTI systems, as shown in Fig. 3.11. Let the impulse responses of the two systems be specified as h1[n] and h2[n], respectively.

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Fig. 3.11 Parallel interconnection of DT systems The output of the system is y[n]. Let z[n] represent the output of the first system with impulse response h1[n]. Let w[n] represent the output of the second system with impulse response h2[n]. The output z[n] can be written as

z[n] =

n

∑ x[m]h [n − m]

m = −∞

(3.132)

1

The symbol * is used to represent the convolution operation. The output w[n] of the second system can be written as w[n] = x[n]*h2[n] w[n] =

229

n

∑ x[m]h [n − m]

m = −∞

CT and DT Systems

z[n] = x[n]*h1[n]

(3.133)

2

The output of the parallel interconnection y[n] can be written as = y[n] x[n]* h1[n] + x[n]* h2 [n] = y[n]

n

∑ x[m]h [n − m] + ∑ x[m]h [n − m]

m = −∞

= y[n]

1

m = −∞

2

(3.134)

n

∑ x[m]{h [n − m] + h [n − m]}

m = −∞

1

2

Let the impulse response of the parallel configuration be written as h[n] given by h= [n] h1[n] + h2 [n]

(3.135)

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= y[n]

n

∑ x[m]{h [n − m] + h [n − m]}

m = −∞

= y[n]

n

1

[n − m] ∑ x[m]h=

2

(3.136) x[n]* h[n]

m = −∞

We can conclude that when the two systems are connected in parallel, the impulse response of the parallel configuration system is the addition of impulse responses of the two systems. Let us go through some numerical problems to illustrate the concepts. Example 3.51

Signals and Systems

Consider the configuration shown in Fig. 3.12 with impulse responses given by h1(t), h2(t), h3(t) for the systems, as shown. Find the impulse response of the overall configuration.

230

Fig. 3.12 Configuration of three systems for Example 3.51 Solution

Firstly we will find the impulse response of the parallel interconnection of two systems with impulse response h1(t) and h2(t). The impulse response of the parallel configuration is h(t) = h1(t) + h2(t). The third system is connected in series with the parallel configuration. Hence, the impulse response of the series configuration will be the convolution of the two impulse responses. We find that the distributive law holds good. h '(t ) = [h1 (t ) + h2 (t )]* h3 (t ) = h1 (t ) * h3 (t ) + h2 (t ) * h3 (t )

(3.137)

Example 3.52

Consider the configuration shown in Fig. 3.13 with impulse responses given by h1(t), h2(t), h3(t) for the systems, as shown in the figure below. Find the impulse response of the overall configuration. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Fig. 3.13 Configuration of four systems for Example 3.52 Solution

Let us find the impulse response of the series system h(t) giving output as z(t) (3.138)

This series configuration is in parallel with h2(t). The impulse response of the parallel configuration is h '(t ) =[h(t ) + h2 (t )] =[h1 (t ) * h3 (t )] + h2 (t )

(3.139)

The system with impulse response h3(t) is connected in series with this parallel configuration. The impulse response of the overall system is then given by h ''(t ) = [h(t ) + h2 (t )]* h3 (t ) = {[h1 (t ) * h3 (t )] + h2 (t )} * h3 (t )

(3.140)

CT and DT Systems

h(t ) = h1 (t ) * h3 (t ).

231

Example 3.53

Consider the configuration shown in Fig. 3.14 with impulse responses given by h1(t), h2(t), h3(t) for the systems, as shown in the figure below. Find the impulse response of the overall configuration.

Fig. 3.14 Configuration of four systems for Example 3.53 Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Solution

Let us find the impulse response of the parallel system h(t) giving output as z(t) h= (t ) h1 (t ) + h2 (t ).

(3.141)

This parallel configuration is in series with h3(t). The impulse response of the series configuration is h= '(t ) [h1 (t ) + h2 (t )]* h3 (t )

(3.142)

The system with impulse response h1(t) is connected in series with this series configuration. The impulse response of the overall system is then given by h= ''(t ) [h1 (t ) + h2 (t )]* h3 (t ) * h1 (t )

(3.143)

Signals and Systems

Example 3.54

232

Consider the overall impulse response of the system given by h ''(t ) = [h1 (t ) * h2 (t )] + [h3 (t ) * h1 (t )] for the three systems with impulse responses h1(t), h2(t), h3(t). Draw the configuration. Solution

The overall impulse response for the configuration indicates that the series combination of h1(t) and h2(t) is connected in parallel with series combination of h3(t) and h1(t). It can be drawn as shown in Fig. 3.15.

Fig. 3.15 Configuration of four systems for Example 3.54 We will now solve some examples for DT systems. Example 3.55

Consider the configuration shown in Fig. 3.16 with impulse responses given by h1[t], h2[t] and h3[t] in Fig. 3.16 for the systems, as shown in the figure below. Find the impulse response of the overall configuration. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Fig. 3.16 Configuration of five systems for Example 3.55 Solution

Let us find the impulse response of the series system h[n] giving output as z[n] (3.144)

Let us find the impulse response of the series system h¢[n] giving output as w[n] h '[n] = h2 [n]* h3 [n].

(3.145)

These series configurations are in parallel. The impulse response of the parallel configuration h’’[n] is h ''[n] =[h[n] + h '(t )] =[h1[n]* h3[n]] + [h2 [n]* h3[n]]

(3.146)

CT and DT Systems

h[n] = h1[n]* h3[n].

233

The system with impulse response h3[n] is connected in series with this parallel configuration. The impulse response of the overall system hoverall is then given by h[n]overall = [h[n] + h '[n]]* h3 [n] = {[h1[n]* h3[n]] + [h2 [n]* h3[n]]} * h3 (t )

(3.147)

Example 3.56

Consider the configuration shown in Fig. 3.17 with impulse responses given by h1[n], h2[n], h3[n] for the systems, as shown in the figure below. Find the impulse response of the overall configuration. Solution

Let us find the impulse response of the parallel system h[n] giving output as z[n] h= [n] h1[n] + h2 [n].

(3.148)

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Fig. 3.17 Configuration of three systems for Example 3.56 This parallel configuration is in series with h3[n]. The impulse response of the series configuration is h= '(t ) [h1 (t ) + h2 (t )]* h3 (t )

(3.149)

Signals and Systems

Example 3.57

234

Consider the overall impulse response of the system given by h¢¢[n] = {[h1[n]* h2[n]]+[h3[n]*h1[n]]}*h1[n] for the three systems with impulse responses h1[n], h2[n], h3[n]. Draw the configuration. Solution

The overall impulse response for the configuration indicates that the series combination of h1[n] and h2[n] is connected in parallel with series combination of h3[n] and h1[n]. This parallel configuration is in series with h1[n]. It can be drawn as shown in Fig. 3.18.

Fig. 3.18 Configuration of five systems for Example 3.57 Things to remember

If the two systems are connected in series, their impulse responses get convolved. If the two systems are connected in parallel, their impulse responses get added. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

Concept Check

• What is the impulse response of the series of two systems? • How will you find the impulse response of the parallel configuration? • Does the rule for series and parallel interconnection hold good for DT systems?

Summary In this chapter, we have described and explained the properties of DT systems. linearity and shift invariance. It was shown that a system is linear when it is homogeneous as well as additive. It was emphasized that if the transfer curve of the system is linear passing through the origin, then it is linear. Linearity has a meaning more than this. If the system obeys the principle of superposition i.e., additivity and homogeneity, then the system is linear. We then defined the property of time/shift invariance. If the system is linear, the input signal can be suitably decomposed into component signals and the corresponding outputs for the component signals one at a time can be calculated by assuming all other inputs equal to zero. The component outputs can be scaled and added to generate the output of the system for the input signal. This is exactly the property of superposition. The system is said to be time/shift invariant if the input to the system is shifted impulse d(n – k), then it results in a shifted impulse response of h(n – k). The linear and shift invariant system is termed as LTI system. If the system is LTI, one can characterize the system in terms of its impulse response. The calculation of the output for any given input in case of LTI system gets greatly simplified due to the principle of superposition.

CT and DT Systems

• We have discussed the important properties of DT systems, namely,

235

• We further concentrated on causality and memory property of systems. We

defined the property of causality for systems. The system is said to be causal if the present output of the system depends only on current and past input or output at previous instant. Causal systems are practically realizable or implementable. The system of a human being is causal as we always keep on learning from the past inputs and past outputs of the system. The future inputs have no effect on our act at current or present time. The present and past inputs have meaning only for temporal systems where time is an independent variable. In case of spatial domain systems, present and past input has no meaning. Non-causal temporal systems can be implemented if some delay is tolerable. We can generate a bench mark for system performance using noncausal systems. Offline systems can always be implemented as non-causal systems. The system is said to have memory or said to be dynamic if its current output depends on previous, future input or previous and future output signals. The system is said to be memoryless or instantaneous if its current output depends only on current input. Examples of memoryless systems are the systems containing only resistive elements.

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• We further discussed invertibility and stability. If it is possible to recover the

input of the system, then the system is said to be invertible. Invertible system also finds applications in communication field. For error-free transmission an equalizer is used at the input of the receiver that has inverse characteristics as that of the channel. Stability is a notion that describes whether the system will be able to follow the input. A system is said to be unstable if its output is out of control or increases without bound. An arbitrary relaxed system (with zero initial conditions) is said to be bounded input bounded output (BIBO) stable if and only if its output is bounded for every bounded input.

• The system can be described as an interconnection of operations. If the time

delay is represented as S block, we can draw the block schematic for the time difference equations. We have discussed series and parallel interconnections of LTI systems and have shown that for series connections, the impulse responses of the individual systems get convolved and for parallel interconnections, the impulse responses of individual systems get added.

Signals and Systems

Multiple Choice Questions 1. The system is linear if

236

2. The system is causal when the current output sample depends on

(a) it is homogeneous (b) bit is additive (c) it is additive or homogeneous (d) it is additive and homogeneous (a) current input sample (b) current or next and past input samples (c) current and/or past input samples and/or past output samples (d) next or past input samples or past output samples 3. The range of values of “a” for which the system with impulse response h(n) = anu(n) is stable is (a) |a| > 1

(b)

(c) a > 0

(d) a < 0

|a| < 1

4. If the transfer graph for a system is linear and passes through origin (a) the system is nonlinear (b) the system s linear (c) the system may be additive (d) the system may be homogeneous 5. The system of human being is (a) non-causal

(b)

non-linear

(c) causal

(d) non-linear and non-causal

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6. The system is said to be memoryless if (a) Only on a current input sample (b) current or next and past input samples (c) current and/or past input samples and/or past output samples (d) next or past input samples or past output samples 7. The following system is invertible (a) different transforms (b) all systems (c) if the equation exists to find signal values (d) non-linear system such as squaring device 8. The bench mark system can be designed using (a) causal systems

(b)

invertible systems

(c) non-invertible systems

(d) non-causal systems

9. The following systems are prone to noise (b)

causal systems

(c) non stable systems

(d) invertible systems

(c) the output goes on increasing

CT and DT Systems

(a) non-causal systems

(d) the output decreases

237

10. The system is BIBO stable if (a) the output is bonded for every bounded input (b) the output is always bounded

11. The system is called as LTI if (a) the system is linear (b) the system is linear and time invariant (c)

the system is time invariant

(d) the system is additive 12. The system given by y(t) = x(t) + 5 (a) is memoryless (b) is with memory (c) is unstable and without memory (d) is stable and with memory 13. The system given by y[n] = x[3 – n] is (a) causal

(b)

anti causal

(c) invertible and causal

(d) causal and non-invertible

14. The system given by y(t) = x2(t) is (a) invertible

(b)

non-invertible

(c) invertible with memory

(d) invertible and memoryless

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15. The system given by y[n] = (n – 1)x[n] is (a) time invariant

(b)

time dependant

(c) time bound

(d) time variant

16. The system given by y[n] = x[n – 1]sin(nω) is (a) time invariant

(b)

LTI

(c) Non-linear and time invariant

(d) Time variant

17. Series interconnection of system results in (a) addition of the impulse responses (b) convolution of impulse responses (c) subtraction of impulse responses (d) multiplication of impulse responses 18. Parallel interconnection of systems results in (a) addition of the impulse responses (b) convolution of impulse responses (c) subtraction of impulse responses Signals and Systems

(d) multiplication of impulse responses

Review Questions 3.1

238

What is linearity? Define additivity and homogeneity. Is the transfer curve for a linear system always linear? Explain physical significance of linearity.

3.2

What is the time/shift invariance property of systems? Explain physical significance of shift invariance property.

3.3

Explain property of superposition.

3.4

When will you say that the system is memoryless? Give one example of a memoryless system.

3.5

Define causality for a system. Can we design and use a non-causal system? Is a causal system a requirement for spatial systems?

3.6

Explain the meaning of causality for a system of a human being.

3.7

What is invertibility? Can we use a non-invertible transform for processing a signal?

3.8

Explain the meaning of BIBO stability for a system.

3.9

Explain the physical significance of stability.

3.10 How will you interpret the system as interconnection of operators? Explain using a suitable example. 3.11 Find the impulse response for a series interconnected and parallel interconnected systems. Prove that the impulse response of the series interconnection of two LTI CT systems is a convolution of the two impulse responses. Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

3.12 Find the impulse response for a series interconnected and parallel interconnected systems. Prove that the impulse response of the series interconnection of two LTI DT systems is a convolution of the two impulse responses.

Problems 3.1

Is the system given by y[n] = x[–n] a linear and shift invariant system?

3.2

Is the system given by y(t) = x(t – 2) a linear and shift invariant system?

3.3

Verify that the systems given by y(n) = x[n]cos(wn) and y[n] = nx[n] are shift variant.

3.4

Check if the systems given by y(t) = (t – 1) x(t) and y(t) = x(t)cos(wt + p /4) are shift invariant?

3.5

Find if the following systems are time invariant. (a) y[n] = x[n] – x[n – 1] (c) y[n] = x[1 – n] (d) y[n] = x[n]sin(wn) (e) y(t) = x(t) + x(t + 1) (f)

y(t) = t2x(t)

(g) y(t) = x(4 – t) (h) y(t) = x(t)sin(t) 3.6

CT and DT Systems

(b) y[n] = nx[n]

239

Find if the following systems are linear. (a) y[n] = (n + 1)x[n] (b) y[n] = x[n2] (c) y[n] = x3[n] (d) y[n] = 2x[n] + 3 (e) y(t) =(t + 2)x(t) (f)

y(t) = x3(t)

(g) y(t) = 3x(t) + 1 (h) y(t) = sin(t)x(t) 3.7

Find if the following systems are causal. (a) y[n] = 5x[n] (b) y[n] =

n +1

∑ x(k)

k = −∞

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(c) y[n] = x[3 – n] (d) y[n] = x[3n] (e) y(t) = x(t2) (f)

y(t) = x(5 – t)

(g) y(t) = x(2t – 2) (h) y(t) = x(–2t) 3.8

Find if the following systems are memoryless (a) y(t) = e–2x(t) (b) y(t) = cos(x(t)) (c) y[n] = 5x[n] + 2x[n]u[n] (d)

t /3

y(t ) = ∫ x(τ )dτ −∞

Signals and Systems

(e) y(t) = x(7 – 2t)

240

(f) 3.9

y(t) = x(t/5)

Find if the following systems are stable. (a) y(t) = cos(x(t)) (b) y[n] = log10 (|x[n]|) (c) y[n] = cos(2p x[n]) + x[n] (d)

y(t ) =

d −t [e x(t )] dt

(e) y(t) = x(t /3) y[n] (f)=

n

∑ x[m + 3]

m = −∞

∞

= (g) y[n] x[n] ∑ δ [n − 5m] m = −∞

3.10 Find if the following systems are invertible. (a)= y[n]

n

∑ x[m + 3]

m −∞

(b) y[n]= x[n − 1] + 4 (c) y(t) = x3(t) (d) y(t ) = x(t / 9) Downloaded from https:/www.cambridge.org/core. The University of Melbourne Libraries, on 13 Jan 2017 at 11:43:09, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781316536483.004

(e)

y(t ) = x(t )

(f)

y[n] = x[2n]

3.11 Represent the following systems in terms of interconnection of operators (a) y(t) = x(t) + x(t – 3) + y(t – 6) (b) y(t) = x(t – 1) –y(t – 2) –y(t – 3) (c) y[n] = x[n] + y[n – 1] + y[n – 2] (d) y[n] = x[n – 2] + y[n – 2] – y[n – 4] 3.12 Find the overall impulse response for the interconnection of three systems.

(b)

CT and DT Systems

(a)

241

(c)

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(d)

Signals and Systems

(e)

242

3.13 Find the possible interconnection for the following equation of the overall impulse response of the system. (a)

hoverall [n] = {[h1[n] + h2 [n]]*[h3 [n] + h1[n]]} * h1[n]

(b) hoverall [n] = {[h1[n]* h2 [n]] + [h3 [n]* h1[n]]} *[h1[n] + h2 [n]] (c)

= hoverall (t ) {h1 (t ) + [h2 (t ) * h3 (t )]} * h3 (t )

Answers Multiple Choice Questions 1 (a)

2 (c)

3 (b)

4 (b)

5 (c)

6 (a)

7 (a)

8 (d)

9 (c)

10 (a)

11 (b)

12 (a)

13 (b)

14 (b)

15 (d)

16 (d)

17 (b)

18 (a)

Problems 3.1

Yes – Linear and shift invariant

3.2

Yes – linear and shift invariant

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3.3

Yes, systems are time variant

3.4

The systems are time variant

3.5 (a) Yes

(b) No

(c) Yes

(f) No

(g) Yes

(h) No

(d) No

(e) Yes

3.6 (a) Yes

(b) Yes

(c) No

(d) Yes

(e) Yes

(f) No

(g) No

(h) No

(a) Yes

(b) No

(c) No

(d) No

(e) No

(f) No

(g) No

(h) No

(a) Yes

(b) Yes

(c) Yes

(d) No

(e) No

(a) Yes

(b) Yes

(c) Yes

(d) Yes

(e) Yes

(f) No

(g) Yes

(a) No

(b) Yes

3.7.

(f) No 3.9

CT and DT Systems

3.8

243

3.10 (c) Yes

(d) Yes

(e) No

(f) No 3.11 (1)

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3.11 (2)

Signals and Systems

3.11 (3)

244

3.11 (4)

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= (t ) {[h1 (t ) + h2 (t )]* h3 (t ) 3.12 (a) hoverall (b) hoverall (t ) = {[h1 (t ) * h2 (t )] + [h2 (t ) * h3 (t )]} * h3 (t ) (c) hoverall (t ) {h1 (t ) + [h2 (t ) * h3 (t )]} * h3 (t ) * h2 (t ) = (d) hoverall = [n] {[h1[n]] + [h3 [n]* h1[n]]} * h1[n] (e) h ''[n] = {[h1[n]* h2 [n]] + [h3 [n]* h1[n]]}

(b)

CT and DT Systems

3.13 (a)

245

(c)

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